Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{4r^2 - 24r - 160}{r^2 - 3r - 70} \times \dfrac{r + 1}{r + 4} $
Explanation: First factor out any common factors. $n = \dfrac{4(r^2 - 6r - 40)}{r^2 - 3r - 70} \times \dfrac{r + 1}{r + 4} $ Then factor the quadratic expressions. $n = \dfrac {4(r - 10)(r + 4)} {(r - 10)(r + 7)} \times \dfrac {r + 1} {r + 4} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac { 4(r - 10)(r + 4) \times (r + 1)} { (r - 10)(r + 7) \times (r + 4)} $ $n = \dfrac {4(r - 10)(r + 4)(r + 1)} {(r - 10)(r + 7)(r + 4)} $ Notice that $(r - 10)$ and $(r + 4)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac {4\cancel{(r - 10)}(r + 4)(r + 1)} {\cancel{(r - 10)}(r + 7)(r + 4)} $ We are dividing by $r - 10$ , so $r - 10 \neq 0$ Therefore, $r \neq 10$ $n = \dfrac {4\cancel{(r - 10)}\cancel{(r + 4)}(r + 1)} {\cancel{(r - 10)}(r + 7)\cancel{(r + 4)}} $ We are dividing by $r + 4$ , so $r + 4 \neq 0$ Therefore, $r \neq -4$ $n = \dfrac {4(r + 1)} {r + 7} $ $ n = \dfrac{4(r + 1)}{r + 7}; r \neq 10; r \neq -4 $